Question

Solve each differential equation by variation of parameters

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}}$

Short answer

$y=k_1{e}^x+k_2{e}^{-2x}+\left(e^{-2x}+e^{-x}\right)\ln \left(1+e^x\right)-e^{-x}$

Step-by-step solution

Step 1: Determine the corresponding homogeneous equation

We have the second order differential equation

$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=\frac{1}{1+e^x}$

as the form

$y^{\text{'}\text{'}}+p\left(x\right)y^{\text{'}}+q\left(x\right)y=f\left(x\right),$

and we have to get its general solution by the following technique:

First, we have to find the solution of the corresponding homogeneous differential equation as

$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=0$ …….. (1)

by assuming that $y=e^{mx}$is a solution for the homogeneous differential equation, then follow the following technique:

Differentiate the assumption with respect to x, then we have

$y^{\text{'}}=m{e}^{mx}$

and

$y^{\text{'}\text{'}}=m^2{e}^{mx}$ …….. (2)

After that, substitute with $y=e^{mx}$and equation (2) into the differential equation (1), then we obtain

$\begin{array}{rcl}{m}^2{e}^{mx}+3m{e}^{mx}+2e^{mx}& =& 0\\ e^{mt}\left(m^2+3m+2\right)& =& 0\end{array}$

Since$e^{mx}$ cannot be equal 0, the we have

$\begin{array}{rcl}{m}^2+3m+2& =& 0\\ (m+1)(m+2)& =& 0\end{array}$

Then the roots are

$m_1=-1\text{and}{m}_2=-2$

$\begin{array}{rcl}{y}_h& =& c_1{y}_1+c_2\\ y_h& =& c_1{e}^{-x}+c_2{e}^{2x}\end{array}$

$y_h=c_1{e}^{-x}+c_2{e}^{2x}$ ……. (a)

Determine the particular solution for  differential equation

Second, now we have to find the particular solution of the given differential equation as the following technique:

Since we have$y_1=e^{-x}$ and $y_2=e^{-2x}$, then we can obtain the wronskian as

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{-x}& e^{-2x}\\ -e^{-x}& -2e^{2x}\end{array}\right|\\ & =& \left(e^{-x}\right)\times \left(-2e^{-2x}\right)-\left(e^{-2x}\right)\times \left(-e^{-x}\right)\\ & =& \left(-2e^{-2x}\right)-\left(-e^{-2x-x}\right)\\ & =& -2e^{3x}+e^{-3x}\\ & =& -e^{-3x}\end{array}$

After that, since we have the right side of the given differential equation $f\left(x\right)=\frac{1}{1+e^x}$, then we can obtain the wronskian $W_1$and$W_2$ as

$\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& e^{-2x}\\ \frac{1}{1+e^x}& -2e^{2x}\end{array}\right|\\ & =& 0-e^{-2x}\times \left(\frac{1}{1+e^x}\right)\\ & =& \frac{-e^{2x}}{1+e^x}\end{array}$

and

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(x\right)\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{-x}& 0\\ -e^x& \frac{1}{1+e^x}\end{array}\right|\\ & =& e^{-x}\times \left(\frac{1}{1+e^x}\right)-0\\ & =& \frac{e^{-x}}{1+e^x}\end{array}$

After we had obtained the wronskian $W_1$and $W_2$, we have to obtain $u{\text{'}}_1$and $u{\text{'}}_2$as the following

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W\left(y_1,y_2\right)}\\ & =& \frac{\frac{-e^{-2x}}{1+e^x}}{-e^{-3x}}\\ & =& \frac{-e^{-2x}\times -e^{3x}}{1+e^x}\\ & =& \frac{e^x}{1+e^x}\end{array}$

and

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{\frac{e^x}{1+e^x}}{-e^{3x}}\\ & =& \frac{e^{-x}\times -e^{3x}}{1+e^x}\\ & =& \frac{-e^{2x}}{1+e^x}\end{array}$

After that, we have to do integration for both$u{\text{'}}_1$and$u{\text{'}}_2$neglecting integration constants to obtain$u_1$and$u_2$as the following

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int \frac{e^x}{1+e^x}dx\\ & =& \ln \left(1+e^x\right)\end{array}$

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dx\\ & =& \int \frac{-e^{2x}}{1+e^x}dx\end{array}$

This integration can be obtained by substitution, we set$v=e^x$, then we have$dv=e^{x}dx$, then we have

$\begin{array}{rcl}{u}_2& =& \int \frac{-e^x\times e^x}{1+e^x}dx\\ & =& \int \frac{-vdv}{1+v}dv\\ & =& -\int \left(\frac{v+1-1}{1+v}\right)dv\\ & =& -\int \left(1-\frac{1}{1+v}\right)dv\\ & =& \int \left(\frac{1}{1+v}-1\right)dv\\ & =& \ln (1+v)-v\\ & =& \ln \left(1+e^x\right)-e^x\end{array}$

After that, using the values of$u_1$and$u_2$ the functions$y_1$and$y_2$ , we can obtain the particular solution as

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(\ln \left(1+e^x\right)\right)\times \left(e^{-x}\right)+\left(\ln \left(1+e^x\right)-e^x\right)\times \left(e^{-2x}\right)\\ & =& e^x\ln \left(1+e^x\right)+e^{-2x}\ln \left(1+e^x\right)-e^{-x}\\ & =& \left(e^{-2x}+e^x\right)\ln \left(1+e^x\right)-e^{-x}\end{array}$

$y_p=\left(e^{-2x}+e^x\right)\ln \left(1+e^x\right)-e^{-x}$ ……… (b)

Determine the general solution for given differential equation

Finally, from equations (a) and (b) we can obtain the general solution of the differential equation $y^{\text{'}\text{'}}+3y^{\text{'}}+2y=\frac{1}{1+e^x}$as

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& c_1{e}^{-x}+c_2{e}^{-2x}+\left(e^{-2x}+e^{-x}\right)\ln \left(1+e^x\right)-e^{-x}\\ & =& \left(c_1-1\right)e^{-x}+c_2{e}^{-2x}+\left(e^{2x}+e^{-x}\right)\ln \left(1+e^x\right)-e^{-x}\\ & =& k_1{e}^x+k_2{e}^{-2x}+\left(e^{-2x}+e^{-x}\right)\ln \left(1+e^x\right)-e^{-x}\end{array}$

where $k_1=c_1-1$and $k_2=c_2$.

$y=k_1{e}^x+k_2{e}^{-2x}+\left(e^{-2x}+e^{-x}\right)\ln \left(1+e^x\right)-e^{-x}$